L3 · Chapter 3

Why a broad thermal spectrum shrinks to one line 100 nm from a hot surface

Sandbox Physics · 2026-10-30

Seen from far away, silicon carbide at 300 K emits a broad infrared spectrum with a piece missing in the middle. Lower a probe to 100 nm and the spectrum of the same material shrinks to almost a single line at $178.7\times10^{12}\,\mathrm{rad\,s^{-1}}$ , more than twenty thousand times taller than the peak of the far-field spectrum. The source has not changed; what has changed is the set of modes that can reach the probe.

Thermal currents: the stronger the absorption, the stronger the fluctuation

Charges inside a material move thermally and produce fluctuating currents. The fluctuation–dissipation theorem says the correlation of these currents is set by one quantity: the imaginary part of the permittivity, $\operatorname{Im}\varepsilon$ , which measures how strongly the material absorbs. Where the material absorbs strongly, the currents fluctuate strongly.

Given the currents, a Green function gives the field they produce on the vacuum side; then one sums over all source points and all in-plane wavevectors. Shchegrov and co-workers write the result as the product of chapter 1: the mean energy of a mode, $\Theta$ , times the local density of modes at the probe. The difference is that this density now depends on height.

That one sentence carries the whole chapter: the spectrum changes with height because the set of modes that can reach the probe changes, not because the energy in each mode changes.

\left\langle j_\alpha(\mathbf r,\omega)\,j_\beta^{*}(\mathbf r',\omega')\right\rangle\;\propto\;\omega\,\varepsilon_0\operatorname{Im}\varepsilon(\omega)\,\Theta(\omega,T)\,\delta_{\alpha\beta}\,\delta(\mathbf r-\mathbf r')\,\delta(\omega-\omega')

The structure of Eqs. (25) to (30) of Joulain et al. Currents exist only inside the material and are uncorrelated between different points, directions and frequencies; the

\Theta

in the prefactor is the one from chapter 1.

Source: K. Joulain · A. V. Shchegrov

The travelling part and the evanescent part

The result has two terms. The first is the travelling waves: $P(\omega)$ is the emissivity averaged over angles, independent of height. A black surface has $P=1$ and an energy density of half the cavity value $u_0$ : only one half-space emits, and the other half is vacuum at 0 K.

The second term is the evanescent waves. Each in-plane wavevector $\beta>k_0$ contributes through the imaginary part of the reflection coefficient, times the decay $e^{-2\kappa z}$ from the surface to the probe. The closer the probe, the larger the $\beta$ that reach it and the more modes take part.

One point needs care when reading the literature. The evanescent term of Eq. (45) of Joulain et al. as printed is twice the limit derived below. We fix the prefactor from Eq. (48) of the same paper: the short-range limit of the next section then reproduces Eq. (48), and the black-surface limit is still $u_0/2$ .

u(z,\omega)=\frac{\Theta\,\omega^{2}}{2\pi^{2}c^{3}}\left[P(\omega)+\int_{k_0}^{\infty}\frac{\beta^{3}\,d\beta}{k_0^{3}\,\kappa}\left(\operatorname{Im}r_s+\operatorname{Im}r_p\right)e^{-2\kappa z}\right]

The form used by the course model, in

\mathrm{J\,s\,m^{-3}}

P

is the emissivity of Eq. (46) of Joulain et al. averaged over solid angle; the prefactor of the evanescent term is fixed by Eq. (48).

Source: K. Joulain

Very close: energy density grows as the inverse cube of height

When $k_0z\ll1$ the integral comes mostly from $\beta\gg k_0$ . There $\kappa\approx\beta$ , the s part vanishes, and the p part approaches the quasi-static value of chapter 2, $2\operatorname{Im}\varepsilon/|\varepsilon+1|^2$ , which no longer depends on $\beta$ and comes out of the integral.

What remains is $\int\beta^2e^{-2\beta z}d\beta=1/(4z^3)$ . Put back, this is Eq. (48) of Joulain et al.: the energy density scales as $z^{-3}$ and peaks sharply at the frequency where $|\varepsilon+1|$ is smallest. Eq. (8) of Shchegrov et al. has the same height law and the same $\operatorname{Im}\varepsilon/|\varepsilon+1|^2$ .

Damping sets how narrow the peak is. Double the damping of SiC and the model’s peak height at 100 nm halves, while the energy integrated over the whole spectrum changes by only 2.5%: width and height trade against each other, and the total is set by the number of modes near the line where $\varepsilon\approx-1$ .

Metals such as gold obey the same limit, but their flat asymptote lies in the near ultraviolet, where the mean mode energy is practically zero, so their near-field spectrum shows no line.

\int_0^{\infty}\beta^{2}e^{-2\beta z}\,d\beta=\frac{1}{4z^{3}}

The only integral needed over the in-plane wavevector.

u(z,\omega)\;\xrightarrow{\;k_0z\ll1\;}\;\frac{\omega^{2}\operatorname{Im}\varepsilon}{4\pi^{2}c^{3}\,|\varepsilon+1|^{2}\,(k_0z)^{3}}\,\Theta(\omega,T),\qquad \frac{u}{u_0}=\frac{\operatorname{Im}\varepsilon}{4\,|\varepsilon+1|^{2}\,(k_0z)^{3}}

Eq. (48) of Joulain et al. The right-hand form is its ratio to the cavity energy density

u_0

, convenient for numbers.

Source: K. Joulain · A. V. Shchegrov

Worked example: how tall the line is at 100 nm

Problem. A probe sits 100 nm above SiC at 300 K. Find the ratio of the energy density at $\omega=178.7\times10^{12}\,\mathrm{rad\,s^{-1}}$ to $u_0$ , decide whether the short-range limit can be trusted here, and estimate how much it grows when the probe drops to 10 nm.

Model. Use the ratio form of the short-range limit, which needs only $\varepsilon$ and $k_0z$ .

Numbers. Chapter 2 gave $\operatorname{Im}\varepsilon/|\varepsilon+1|^2=7.697$ . $k_0=178.7\times10^{12}/(2.998\times10^{8})=5.961\times10^{5}\,\mathrm{m^{-1}}$ , so $k_0z=0.05961$ and $(k_0z)^3=2.118\times10^{-4}$ . Then $u/u_0=7.697/(4\times2.118\times10^{-4})=9.09\times10^{3}$ .

Check. The full model gives $9.16\times10^{3}$ , 0.8% above the limit; at 10 nm the two differ by 0.02%. Away from the line, at 166, the full value at 100 nm is 14% above the limit, and at 210, where there is no surface wave, it is 2.7 times larger. The limit can be trusted only near the line and close enough to the surface.

From 100 nm down to 10 nm, $z^{-3}$ predicts a factor of 1000 and the full model gives 992. At 100 nm, 99.55% of the energy reaching the probe is carried by evanescent waves, and the line is $2.1\times10^{4}$ times taller than the peak of the far-field spectrum. Describing the same height, Joulain and co-workers write that the emission is almost monochromatic around $1.78\times10^{14}\,\mathrm{rad\,s^{-1}}$ and that the energy density has grown by more than four orders of magnitude.

Source: K. Joulain · A. V. Shchegrov

Try it in the Lab

Go all the way down to 100 nm

Predict first: as the probe goes from 1 mm to 100 nm, where does the strongest frequency move? Will there be two peaks on the way?
Choose the Approach scan preset and follow the strongest frequency and the Energy in bound waves readout as the height changes. Stop near 2 µm and check whether a taller peak still sits on the low-frequency side.
Choose the Almost one color preset, read the peak over the far-field peak and compare with the worked example. Open More numbers and read the ratio of the exact value to the short-range limit at 166.

What damping and temperature each change

Predict first: if the damping doubles, does the line get taller, shorter or wider? What if the sample goes from 300 K to 600 K?
In the Almost one color preset, set the damping multiplier to 2 and back to 1. Then choose the Hotter sample preset.
Explain the effect of damping with the roughly constant integrated energy, and the effect of temperature with the ratio of $\Theta$ found in chapter 1, 10.7.

Exercises with explained answers

In the short-range limit, what is $u/u_0$ for SiC at 166 at a height of 100 nm? How does it compare with 178.7?
Show the answer
At 166, $\operatorname{Im}\varepsilon/|\varepsilon+1|^2=0.407/42.2=9.65\times10^{-3}$ , $k_0z=5.537\times10^{5}\times10^{-7}=0.0554$ , $(k_0z)^3=1.70\times10^{-4}$ , and $u/u_0=9.65\times10^{-3}/(4\times1.70\times10^{-4})=14.2$ . At 178.7 it is $9.09\times10^{3}$ , about 640 times more. The two frequencies also have different $\Theta$ , but here each value is a ratio to $u_0$ at its own frequency, so the difference comes entirely from $|\varepsilon+1|$ .
Why is the energy density above a black surface only half the cavity value? Does it also grow as the inverse cube of height close to a black surface?
Show the answer
A black surface emits from the half-space below only; above it is vacuum at 0 K, with no radiation coming back, so the result is half. A black surface reflects nothing, $r=0$ in the model, so the evanescent term vanishes and the energy density is $u_0/2$ at every height, with no near-field growth. Near-field growth needs a real material that absorbs and can hold an evanescent field at its surface.
At 1 µm the model puts the peak at 178.2 rather than 178.7 (in $10^{12}\,\mathrm{rad\,s^{-1}}$ ). Why does the peak sit slightly lower when the probe is farther away?
Show the answer
Each evanescent wave reaching the probe carries the factor $e^{-2\kappa z}$ . At 1 µm, waves with a decay constant above about $1/(2z)=5\times10^{5}\,\mathrm{m^{-1}}$ are already weak. The dispersion curve of chapter 2 shows that the closer a surface wave is to the flat asymptote, the larger its wavevector and decay constant. The farther the probe, the more of exactly those modes are suppressed; what remains favours the lower part of the dispersion curve, and the peak moves slightly down. At 100 nm it is back at 178.7.
The sample goes from 300 K to 600 K with the probe at 100 nm. By what factor does the peak frequency of the far-field spectrum change? Does the line move? By what factor does its height change?
Show the answer
The far-field peak frequency is proportional to temperature, so it roughly doubles. The line position is set by $\varepsilon=-1$ , and the model permittivity does not depend on temperature, so it stays put. The line height is proportional to $\Theta(178.7,T)$ , which is 10.7 times larger at 600 K.

Sources and what we recomputed

A. V. Shchegrov, K. Joulain, R. Carminati and J.-J. Greffet, Near-field spectral effects due to electromagnetic surface excitations, Physical Review Letters 85, 1548–1551 (2000), HAL hal-01624761v1
Full text of the published letter in the HAL archive. Eq. (1): the Planck spectrum as mode energy times mode density. Eqs. (2) to (5): thermal currents from the fluctuation–dissipation theorem and the energy density of the emitted electric field. Eq. (6): the surface-wave dispersion relation. Eqs. (7) and (8): the near-field spectrum as mode energy times a local mode density that grows as one over height cubed. Parameters of SiC on p. 1549. Fig. 1: spectra at 1000, 2 and 0.1 µm above SiC at 300 K, with the line at 178.7 in units of 10 to the 12 rad/s. Fig. 3: decay with height at 166 and 210 in the same units. We recomputed the line position, the height law and the decay at 166; we did not digitise the figures.
K. Joulain, J.-P. Mulet, F. Marquier, R. Carminati and J.-J. Greffet, Surface electromagnetic waves thermally excited: radiative heat transfer, coherence properties and Casimir forces revisited in the near field, arXiv:physics/0504068v2; Surface Science Reports 57, 59–112 (2005)
Version 2, 24 April 2005. Sec. 2: surface waves, the Drude permittivity (Eq. 21) and the SiC oscillator (Eq. 22). Sec. 3.2: the fluctuation–dissipation theorem and the mean oscillator energy (Eqs. 25 to 30). Secs. 4.1 and 4.2: the emitted energy density (Eqs. 44 to 48), emissivity as the half-sum of the two transmission factors and its link to Kirchhoff’s law, and Fig. 6 at 100 µm, 1 µm and 100 nm above SiC. Sec. 7: heat flux between two half-spaces (Eqs. 89 to 93), the growth as the inverse square of the gap, and Figs. 17 and 18, which they reproduce from another paper. Sec. 8: the Casimir force from the same fluctuating fields. The evanescent term of Eq. 45 as printed carries an extra factor of two; we fixed the prefactor from Eq. 48. The prefactor of Eq. 93 is garbled in the text version we read, so we quote only its inverse-square dependence on the gap.